Question: The matrix for reflecting over a certain line $\ell,$ which passes through the origin, is given by
\[\begin{pmatrix} \frac{7}{25} & -\frac{24}{25} \\ -\frac{24}{25} & -\frac{7}{25} \end{pmatrix}.\]Find the direction vector of line $\ell.$  Enter your answer in the form $\begin{pmatrix} a \\ b \end{pmatrix},$ where $a,$ and $b$ are integers, $a > 0,$ and $\gcd(|a|,|b|) = 1.$
Since $\begin{pmatrix} a \\ b \end{pmatrix}$ actually lies on $\ell,$ the reflection takes this vector to itself.

[asy]
unitsize(1.5 cm);

pair D = (4,-3), V = (2,1), P = (V + reflect((0,0),D)*(V))/2;

draw((4,-3)/2--(-4,3)/2,dashed);
draw((-2,0)--(2,0));
draw((0,-2)--(0,2));
draw((0,0)--P,Arrow(6));

label("$\ell$", (4,-3)/2, SE);
[/asy]

Then
\[\begin{pmatrix} \frac{7}{25} & -\frac{24}{25} \\ -\frac{24}{25} & -\frac{7}{25} \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} a \\ b \end{pmatrix}.\]This gives us
\[\begin{pmatrix} \frac{7}{25} a - \frac{24}{25} b \\ -\frac{24}{25} a - \frac{7}{25} b \end{pmatrix} = \begin{pmatrix} a \\ b \end{pmatrix}.\]Then $\frac{7}{25} a - \frac{24}{25} b = a$ and $-\frac{24}{25} a - \frac{7}{25} b = b.$  Either equation reduces to $b = -\frac{3}{4} a,$ so the vector we seek is $\boxed{\begin{pmatrix} 4 \\ -3 \end{pmatrix}}.$